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Intern
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A school assigns students to small classrooms in such a way [#permalink]
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20 Apr 2006, 05:46
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A school assigns students to small classrooms in such a way that some of the classrooms can be empty and more than one student can be assigned to a classroom.
Question 1)In how many ways can the school assign 3 students to 2 different small classrooms?
Question 2)In how many ways can the school assign 4 students to 3 different small classrooms?



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Joined: 24 Jan 2006
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Question 1)In how many ways can the school assign 3 students to 2 different small classrooms?
2[(3C3 * 3C0) + (3C2 + 1)]
2(1+4)
10
Question 2)In how many ways can the school assign 4 students to 3 different small classrooms?
2[(4C2 * 4C2) + (4C1 * 4C3) + (4C4 * 4C0)]
2(36 + 16 + 1)
2(53)
106



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Joined: 09 Mar 2006
Posts: 444

1. 8 ( 2^3 )
2. 81 ( 3^4 )
Each student can be assigned to one of X classrooms, so if there are
Y students they can be assigned in X^Y ways.



Intern
Joined: 23 Dec 2005
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Question 1:
Let A,B,C be students and C1, C2 be class rooms
C1 C2
0 ABC
A BC
B AC
C AB
AB C
BC A
CA B
ABC 0
Total possible ways = 8. However, can someone please explain this in a better way rather than listing all the posibilities. As this approach for Question 2 becomes quite cumbersome. Also, please explain in detail...
Last edited by Ethan on 20 Apr 2006, 07:03, edited 1 time in total.



Intern
Joined: 23 Dec 2005
Posts: 18

Deowl,
Have we taken into consideration that there is a possiblity of no student being assigned to a class room. What I am not sure at this point is, does n^m take care of possibility of students not assigned to a classroom?
Thanks in advance.



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Joined: 09 Mar 2006
Posts: 444

Absolutely. Since each student has equal probability to be assigned to any of the classes, all students could be assigned to one class so other classes remain empty.



Intern
Joined: 23 Dec 2005
Posts: 18

The OA for Question 2 is 36 and I digged this question out from some archives. And I have no clue how it can be 36...Any thoughts....



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Joined: 09 Mar 2006
Posts: 444

Ethan
I think you have omitted one important condition from the second question.
( actually this makes sense since who would ask the same question with such a minor modification twice )
The condition is that no rooms should remain empty. In this case we resolve it
in the following way:
1. # of possibilities for room with 2 students: 3
2. # of possibilities to find that lucky couple: 6
3. # of arrangements for remaining students: 2
Total: 3 * 6 * 2 = 36










